Class
DatabaseRow
Properties
Name |
Type |
Read-Only |
Shared |
---|---|---|---|
Description
Used to create new Database rows (records). The methods are used to populate the columns in a row (record).
Property descriptions
DatabaseRow.LastColumnIndex
LastColumnIndex As Integer
The index of the last column in the DatabaseRow.
This property is read-only.
Method descriptions
DatabaseRow.Column
Column(name As String) As DatabaseColumn
Gets the DatabaseColumn for column name.
Column(name As String, Assigns value as Variant)
Sets a value for column name. This is more optimized than using DatabaseColumn.
Creates a new row in the team table:
Var row As New DatabaseRow
row.Column("Name").StringValue = "Penguins"
row.Column("Coach").StringValue = "Bob Roberts"
row.Column("City").StringValue = "Boston"
Try
myDB.AddRow("Team", row)
Catch error As DatabaseException
MessageBox("DB Error: " + error.Message)
End Try
Creates a new row in the team table:
Var row As New DatabaseRow
row.Column("Name") = "Penguins"
row.Column("Coach") = "Bob Roberts"
row.Column("City") = "Boston"
Try
myDB.AddRow("Team", row)
Catch error As DatabaseException
MessageBox("DB Error: " + error.Message)
End Try
DatabaseRow.ColumnAt
ColumnAt(index As Integer) As DatabaseColumn
Returns the DatabaseColumn for the column at position Index.
Get the string value of a column in a DatabaseRow:
' row is a DatabaseRow
Var productName As String
productName = row.ColumnAt(0).StringValue
DatabaseRow.ColumnCount
ColumnCount As Integer
Returns the number of columns in the DatabaseRow.
Sets the number of columns in a listbox (CustomersList) to the number of columns in a database row (row):
CustomersList.ColumnCount = row.ColumnCount
DatabaseRow.Columns
Columns As Iterable
Allows iterating through all the columns in the DatabaseRow with For...Each.
Sample code
Creates a new row in the team table:
Var row As New DatabaseRow
row.Column("Name").StringValue = "Penguins"
row.Column("Coach").StringValue = "Bob Roberts"
row.Column("City").StringValue = "Boston"
row.Column("StartTime").DateTimeValue = New DateTime(1, 1, 1, 6, 0, 0) ' 6AM - no specific date
Try
myDB.AddRow("Team", row)
Catch error As DatabaseException
MessageBox("DB Error: " + error.Message)
End Try
Compatibility
All project types on all supported operating systems.
See also
Object parent class; Database, DatabaseColumn, MySQLCommunityServer, ODBCDatabase, PostgreSQLDatabase, RowSet classes.